What is the mark of the winning team? $10$ teams have participated a competition with five contestants.  According to the results, we give each person a grade. The grades are between $1$ and $50$ and we cannot use a grade twice.  The winner is the team that gets the minimum score.  What scores are possible to get by the winning team.
My attempt:  There is a lower bound $1+2+3+4+5=15$ and also an upper bound $\big\lfloor{\frac{1+2+\dots 50}{10}}\big\rfloor =127$, but I am not sure if we can reach the numbers between them.
 A: 
Proposition: If the winning team can get a score of $n$, and $n>15$, then they can get a score of $n-1$.

Proof: If the score $n$ of the winning team isn't the minimum of $15$, then there must be some player on the winning team who got exactly one point more than some player not on the winning team. Swap the grades of the two, and the winning team gets one point less, while some other team gets one point more, and the other right teams are unchanged. Thus the winning team is still the winning team, with $n-1$ points.
This proves that the winning team can "reach the numbers between them". What's left for you is finding the actual maximal score they can get. In other words, can they get $127$ and still be the winning team?
A: It's possible to achieve $127$, so using Arthur's answer we can do all values from $15$ to $127$.
To do it, start by dividing $\{31,...,50\}$ into ten pairs that each sum to $81$. Give each team one of these pairs. This reduces the problem to dividing $\{1,...,30\}$ into ten triples such that the sum of each triple is $46$ or $47$.
One such triple is $1+20+26=47$. Increase the first and third numbers by $1$ and decrease the middle by $1$ to get another triple with the same sum, and repeat this process until you have five triples with sum $47$ using the numbers $1$-$5$, $26$-$30$ and evens from $12$-$20$. Now do the same starting from $6+19+21=46$ to use up the remaining numbers.
A: Every number from $15$ up to $127$ (in fact, up to $240$) can be written as sum of five distinct numbers $\in\{1,2,\ldots,50\}$. Indeed, this is evidently possible for $15=1+2+3+4+5$. Assume it is possible for $n$ with $15\le n<240$, say $n=a_1+a_2+a_3+a_4+a_5$ with $1\le a_1<a_2<a_3<a_4<a_5\le 50$.
If $a_i+1<a_{i+1}$ for some $i$, $1\le i<5$, then we are allowed to replace $a_i$ with $a_i+1$ and obtain a representation for $n+1$. The same holds if $a_5<50$. Thus we are left only with the case that $a_{i+1}=a_i+1$ for $1\le i<5$ and $a_5=50$, but then $n=46+47+48+49+50=240$, contradicting our assumption. We conclude that $n+1$ can also be written that way.
